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1916]                                MEMBRANES, BATiS,  AND PLATES                                    42*7
We proceed then to examine the solution for a plate definitely terminated at distances I, and there supported. For this purpose we resume the general solution (6),
w=ainny {(A + Boc} e~nx + (C + Lx) enx],...............(11)
which already satisfies the conditions of a supported edge at y = 0, y = TT. At x = Q, the condition is as before dw/dx = 0. At x=l the conditions for a supported edge give first w  0, and therefore tfwjdf = 0. The second condition then reduces to d-wjdx* = 0. Applying these conditions to (11) we find
D = Be~*nl,    C = - e-2nl (A + 21B)................(12)
It remains to introduce the condition to be satisfied at x = 0.    In general
~ = sin ny [e~nx {- n (A + Bx) + B}+ e {n (C + Dx) + }];.. .(13)
and since this is to vanish when x = 0,
~nA + B + nC + D=Q.........................(14)
By means of (12), (14) A, C, D may be expressed in terms of B, and we find
^ = ^T^fle~nx \~ *> + (*l~ ^ e~Znl] + e~n(Zl~x) {~ W*- ) + a*-2"1}]- (15)
In (15) the square bracket is negative for any value of* between 0 and I, for it may be written in the form
-xe~nx {1 - e-2n(2z-*)} _ (21  x}e~lnl {enx - e~nx}..........(16)
When x  0 it vanishes, and when x =1 it becomes
-2le-2nl(enl  e-nl).
It appears then that for any fixed value of y there is no change in the sign of dwjdoo over the whole range from  = 0 to x  l. And when n = I, this sign does not alter with y. As to the sign of w when x = 0, we have then from (11)
, 4       ^          .          e2n?_g-2nZ
w = sin ny(A + 0) = E sin ny ~-^JT+
so that dw/doc in (15) has throughout the opposite sign to that of the initial value of w. And since w = 0 when oa = I, it follows that for every value of y the sign of w remains unchanged from CD = 0 to x = I. Further, if n = 1, this sign is the same whatever be the value of y. Every point in the plate is deflected in the same direction.
Let us now suppose that the plate is clamped at x=  I, instead of merely supported. The conditions are of course w = 0, dw/dx= 0. They give
I)},   .....................(IT)
.The condition at ac = 0 is that already expressed in (14).
* [The factor enl has been omitted from the denominator; with Z=oo the corrected result agrees with (7) when x = Q, if B = nA.   W. F. S.] side of the rectangle is relatively long. It seems therefore desirable to inquire more closely into this question.